Number Equation Matrix

Question

Can somebody please solve this? My daughter's school teacher gave her this puzzle to solve at home. But to me it seems a little out of order, and that's why I am asking here for help.

Answer 1

I think the answer is as folows

Reasoning

Usually, in these types of puzzles (when there are nine slots) there is an extra restriction of filling in each digit from 1-9 exactly once. With that in mind, the middle column must automatically have both 2 and 5. Then, for the top row to work, we must have the 2 at the top and the 5 at the bottom and, assuming we perform row operations left to right, it must be that we have 3 + 2 x 4 = (3 + 2) x 4 = 20 in the top row.
From there, the middle row must contain 1 and 6 and if 1 is on the left then 4 must be in the lower left corner. But, since we've already used 4, it must be that 6 is on the left and 1 on the right. The rest of the grid then follows quickly.

Answer 2

Assuming the order of operation is preserved (this is significant only for the first line) and all numbers must be integers

there are 13 different solutions (however, all of them but one involve at least one negative number), as follows (numbers are from left to right, from top to bottom):

 5       1       15      -2      -3      3       10      -8
 -5      -1      -25     -2      -3      -7      -10     32
 10      2       5       -2      -3      8       5       2
 -10     -2      -15     -2      -3      -12     -5      22
 25      5       -1      -2      -3      23      2       8
 -25     -5      -9      -2      -3      -27     -2      16
 50      10      -3      -2      -3      48      1       10
 0       1       20      3       2       3       10      -8
 50      10      -3      3       2       53      1       15
 -4      -1      -24     -3      -2      -7      -10     32
 -50     -10     -7      -3      -2      -53     -1      15
 2       2       9       6       1       8       5       2
 -6      -2      -13     -6      -1      -12     -5      22
 

So, the only solution with all numbers positive is

2 + 2 x 9 = 20, 6 x 8 x 1 = 48, 8 x 5 - 2 = 38, 2 + 6 - 8 = 0, 2 x 8 x 5 = 40, 9 - 1 + 2 = 10.

Explanation:

Let A to H be the 8 unknown numbers. We get the following system of equations: $A+BC=20$, $8DE=48$ (or $DE=6$), $FG-H=38$, $A+D-F=0$, $8BG=80$ (or $BG=10$), $C-E+H=10$. Now it's clear that $D$ and $E$ must be either 1 and 6, or 2 and 3 in any order (or the opposite numbers -1 and -6, or -2 and -3). The same can be said about $B$ and $G$ (1 and 10 or 2 and 5, etc.). So we get $8\times8=64$ different possibilities for $B,D,E,G$, so all of them can be quickly bruteforced. Note that $$F=A+D=20-BC+D=20-B(10+E-H)+D=20-B(10+E-FG+38)+D=20-B(E-FG+48)+D.$$
After simplification, we get $F=(-\frac{20+D}{B}+E+48)/(G-\frac1B)$. So if $F$ is integer, we can quickly compute $A$, $C$ and $H$, which must be all integers.
I've written simple Python script which does the job (giving all 13 solutions): Try it online!

Answer 3

Assuming order of operations is not perserved (multiplication and addition occur from left to right, top to bottom in the column/row), the solution is:

 1 + 1  x 10|20
 +   x    - |
 3 x 8  x 2 |48
 -   x    + |
 4 x 10 - 2 |38
 -----------+
 0   80   10